Using the method of partial fractions, determine the following integral: integral ((8x - 23)/(x^2 - 5x + 4))dx?

(8x-23)/(x²-5x+4) = (8x-23)/(x-1)(x-4)

(8x-23)/(x-1)(x-4) = A/(x-1) + B/(x-4)

(8x-23)/(x-1)(x-4) = [A(x-4) + B(x-1)]/(x-1)(x-4)

(8x-23)/(x-1)(x-4) = [(A+B)x-(4A+B)]/(x-1)(x-4)

==> 8x-23 = (A+B)x -(4A+B)

Matching the coefficients, you get

A+B = 8

4A+B = 23

Solving these simultaneously, you find

A = 5, B = 3

So

(8x-23)/(x²-5x+4) = 5/(x-1)+ 3/(x-4)

∫(8x-23)dx/(x²-5x+4)=∫[5/(x-1)+ 3/(x-4)] dx

= 5∫dx/(x-1) + 3∫dx/(x-4)

= 5ln|x-1| + 3ln|x-4| + C

First factor the denominator:

x^2-5x +4 = (x-4)(x-1)

Now set up the summation of each part:

A/(x-4) +B/(x-1) = ((8x - 23)/(x^2 - 5x + 4))

If you combine the denominators on the left hand side (LHS) you get:

(A(x-1) +B(x-4))/(x^2 - 5x + 4) = ((8x - 23)/(x^2 - 5x + 4))

Now you can cancel the denominators and expand the LHS to get:

Ax -A + Bx -4B = 8x-23

or rearranging:

(A+B)x + (-A - 4B) = 8x-23

Now you have 2 equations to solve for A and B.

A +B = 8,

-A-4B = -23

Solve the first for A

A=8-B

Plug into the second to solve for B

-(8 -B) -4B = -23

-3B = -15

B=5

Plugging back into the first,

A=3

Now you can plug 3 and 5 in above for A and B to get

3/(x-4) + 5/(x-1) = ((8x - 23)/(x^2 - 5x + 4))

and you have your new integral, the solution to which is

3 ln(x-4) + 5 ln(x-1)

(x^2-1)/ (9x^2+6x+5)^2 = A/(9x^2+6x+5) + Bx+C/(9x^2+6x+5)^2 Solving you will get A= 1/9 , B= -2/3 and C= -14/9 -1/54 (9x+1)/(9x^2+6x+5) -1/108 arctan ((3x+1)/2)+C